Problem 77 - Algebra and Trigonometry with Analytic Geometry

Question

Factor the polynomial. $$ 64 x^{3}+27 $$

Step-by-Step Solution

Answer

The factorization is \((4x + 3)(16x^2 - 12x + 9)\).

Step 1: Recognize the Form

The given polynomial is in the form of a sum of cubes: \[ 64x^3 + 27 = (4x)^3 + 3^3 \] This indicates that we can use the sum of cubes formula to factor it.

Step 2: Recall the Sum of Cubes Formula

The sum of cubes formula is: \[ a^3 + b^3 = (a+b)(a^2-ab+b^2) \] In our case, denote \( a = 4x \) and \( b = 3 \).

Step 3: Apply the Formula

Substitute \( a = 4x \) and \( b = 3 \) into the sum of cubes formula: \[ (4x)^3 + 3^3 = (4x + 3)((4x)^2 - (4x)(3) + 3^2) \] This can be simplified to: \[ (4x + 3)(16x^2 - 12x + 9) \]

Step 4: Verify the Result

Expand the factored expression to ensure it equals the original polynomial: First, expand \( (4x + 3)(16x^2 - 12x + 9) \):\[ 4x(16x^2) + 4x(-12x) + 4x(9) + 3(16x^2) + 3(-12x) + 3(9) \] This simplifies to: \[ 64x^3 - 48x^2 + 36x + 48x^2 - 36x + 27 \] Which simplifies further to: \[ 64x^3 + 27 \]This confirms our factorization is correct.

Key Concepts

Sum of Cubes

A sum of cubes is an algebraic expression that takes the form \( a^3 + b^3 \). This structure appears frequently in algebra and can be factored using a specific formula. Understanding this formula is key to simplifying such expressions.

The formula used to factor a sum of cubes is:

• \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \)

In the given exercise, the polynomial \( 64x^3 + 27 \) can be identified as a sum of cubes:

• \( 64x^3 = (4x)^3 \)
• \( 27 = 3^3 \)

This allows us to set \( a = 4x \) and \( b = 3 \), then substitute into the sum of cubes formula.

By substituting \( a \) and \( b \) into the formula, the expression becomes \( (4x + 3)(16x^2 - 12x + 9) \). This breaks the polynomial into simpler factors, making it easier to work with in further algebraic operations. Using this method is a powerful way to handle cubic expressions efficiently.

Polynomials

Polynomials are expressions made up of variables and coefficients, involving addition, subtraction, multiplication, and non-negative integer exponents. They form a fundamental part of algebra.

A polynomial can be expressed in the form \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where \( a_n \) are coefficients and \( x \) is a variable. The highest power of \( x \) determines the degree of the polynomial.

In the original exercise, \( 64x^3 + 27 \) is a polynomial of degree 3, with terms \( 64x^3 \) and \( 27 \). Understanding the structure of polynomials allows us to apply specific factorization techniques like the sum of cubes. Factorization is crucial here because it simplifies complex expressions, making them easier to compute or evaluate.

Recognizing patterns within polynomials, such as common factors or special forms like sums of cubes, is an essential skill in algebra. These patterns let us transform polynomials into more usable forms, aiding in solving equations and simplifying mathematical models.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and operations. They do not have equality signs, unlike equations.

Expressions like \( 64x^3 + 27 \) involve combining terms with different degrees and coefficients. The goal is often to simplify these expressions to make them more manageable. Simplification often entails factorization, where expressions are broken down into products of simpler factors.

Understanding how to manipulate algebraic expressions involves knowing various operations, such as:

• Expanding – Distributing terms over addition
• Simplifying – Reducing expressions by combining like terms
• Factoring – Expressing as a product of factors

In the exercise, the algebraic expression \( 64x^3 + 27 \) was simplified using the sum of cubes factorization technique. This transformed the original expression into the product \((4x + 3)(16x^2 - 12x + 9)\). This transformation makes further computations or evaluations more straightforward and can illuminate properties that aren't immediately apparent in the original form.